Page 169 - DMTH502_LINEAR_ALGEBRA

Unit 12: Introduction and Characteristic Values of Elementary Canonical Forms

is some linear combination of the vectors in . From what we just did, we know that = 0 for Notes
i i
each i. Since each is linearly independent, we see that we have only the trivial linear relation
i
between the vectors in .
Theorem 2: Let T be a linear operator on a finite-dimensional space V. Let c , ..., c be the distinct
1 k
characteristic values of T and let W be the null space of (T – c I). The following are equivalent:
i i
(i) T is diagonalizable
(ii) The characteristic polynomial for T is
dk
di
f = (x – c ) ... (x – c ) )
i k
and dim W = di, i = 1, ... k.
i
(iii) dim W + ... + dim W = dim V.
i k
Proof: We have observed that (i) implies (ii). If the characteristic polynomial f is the product of
linear factors, as in (ii), then d + .. + d = dim V. For, the sum of the d ’s is the degree of the
i k i
characteristic polynomial, and that degree is dim V. Therefore (ii) implies (iii). Suppose (iii)
holds. By the lemma, we must have V = W + ... + W , i.e., the characteristic vectors of T span V.
1 k
The matrix analogue of Theorem 2 may be formulated as follows. Let A be an n × n matrix with
entries in a field F, and let c , ... c be the distinct characteristic values of A in F. For each i, let W
1 k i
be the space of column matrices X (with entries in F) such that
(A – c I)X = 0,
i
and let be an ordered basis for W . The bases , ..., collectively string together to form the
i i 1 k
sequence of columns of a matrix P:
P = [P , P , ...] = ( , ..., )
1 2 1 k
The matrix A is similar over F to a diagonal matrix if and only if P is a square matrix. When P is
square, P is invertible and P AP is diagonal.
–1

3
Example 6: Let T be the linear operator on R which is represented in the standard
ordered basis by the matrix.
5 6 6
A 1 4 2
3 6 4

Let us indicate how one might compute the characteristic polynomial, using various row and
column operations:

x 5 0 6
x 5 6 6
1 x 2 2
1 x 4 2 =
3 6 x 4 3 2 x x 4
x 5 0 6
= (x 2) 1 1 2
3 1 x 4

x 5 0 6
= (x 2) 1 1 2
2 0 x 2

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